Question

Ten points in the plane are given, with no three collinear. Four distinct segments joining pairs of these points are chosen at random, all such segments being equally likely. The probability that some three of the segments form a triangle whose vertices are among the ten given points is $\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

Answer

Solution:

There are $\binom{10}{2} = 45$ segments, so $\binom{45}{4} = 148995$ equally likely choices. Two distinct triangles share at most one edge, so together they use at least $5$ segments; hence a set of $4$ segments contains at most one triangle, and the favorable sets are counted exactly once by choosing a triangle and then a fourth segment: $\binom{10}{3} \cdot (45 - 3) = 120 \cdot 42 = 5040.$

The probability is $\frac{5040}{148995} = \frac{16}{473},$ already in lowest terms ( $473 = 11 \cdot 43$ ), so $m + n = 16 + 473 = 489.$

1999 AIME Problem 10

Solution:

Problem 10 in Other Years